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Proof step 4 from proof alignup-le (previous step)

(m0 % n = 0 ⊃ m0;n * m0 / n + n) < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 ≤ m0 ⊦ n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
m0 % n = 0, m0 < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 ≤ m0 ⊦ n = 0
Proof step 15:
n * m0 / n + n < (m1 % n = 0 ⊃ m1;n * m1 / n + n), m1 ≤ m0 ⊦ m0 % n = 0, n = 0

Rule argument

pos: left 1

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